Find the Value of 2abcosC: Law of Cosines Steps, Proof, and Worked Examples

Set the scene. Station $A$ and station $B$ stand on opposite sides of a mountain — no tape measure will ever cross that peak. The find the value of 2abcosc arithmetic never needed the tape anyway.

The surveyor picks a point $C$ visible from both stations and walks only the reachable legs: $CA = 11.5$ miles and $CB = 9.4$ miles. One more reading, the angle where the legs meet: $\angle C = 59.5^\circ$. The find the value of 2abcosc machinery will need nothing else.

Three facts in hand — two sides and their included angle, all any find the value of 2abcosc computation ever needs. One target: the unreachable $AB$, waiting on one find the value of 2abcosc step.

Two sides and the included angle pin the third — that is where the law of cosines earns its keep. It is also where the find the value of 2abcosc term does its one job. The problem comes from Granville's 1908 trigonometry classic. Example 3 returns to finish the measurement, one find the value of 2abcosc step at a time. The same arithmetic answers every find the value of 2abcosc question you will meet.

Build the formula from one small example you can check in your head. The same arithmetic later powers the find the value of 2abcosc move.

Step 1 — lay out numbers. Take sides $a = 8$ and $b = 5$ with included angle $C = 60^\circ$; ask how long the third side $c$ runs. Small numbers keep the find the value of 2abcosc arithmetic visible.

Step 2 — say it in words. In a right triangle, the hypotenuse's square equals the two legs' squares summed. Let the angle drift off $90^\circ$ and a correction joins: third side's square = the two squares summed − 2 × the product of those sides × the cosine of their angle.

Step 3 — swap in letters. That sentence, written in symbols, is the law of cosines:

$c^2 = a^2 + b^2 - 2ab\cos C$

Step 4 — substitute and compute. $c^2 = 8^2 + 5^2 - 2\times 8\times 5\times\cos 60^\circ = 64 + 25 - 40 = 49$, so $c = 7$. That middle subtraction is the engine inside every find the value of 2abcosc shortcut.

Each letter runs one errand, and every find the value of 2abcosc prompt leans on the last row:

Granville's 1908 text words it precisely. In any triangle, the square of any side equals the sum of the squares of the other two sides. Then subtract twice their product times the cosine of the included angle. Modern worksheets quote it, then ask you to find the value of 2abcosc.

Rotate the letters and all three versions appear — pick the one whose left side is the side you want. A find the value of 2abcosc request always wants the third:

$a^2 = b^2 + c^2 - 2bc\cos A,\qquad b^2 = c^2 + a^2 - 2ca\cos B,\qquad c^2 = a^2 + b^2 - 2ab\cos C$

The trio is one shape with the letters cycled; some older prefaces title the set the laws of cosine. Shorten the hunt to law of cos, stretch it to law of cosines equation — the destination never moves. A worksheet that says find the value of 2abcosc leans on the third version. So does every SAS cousin of a find the value of 2abcosc prompt, as Example 2 will show.

Stay with the same triangle — sides 5 and 8, angle $60^\circ$. Drop a perpendicular $AD$ from $A$ to the opposite side. Watching the correction term appear here makes find the value of 2abcosc feel concrete later.

The altitude creates two right triangles, and the pieces read off at a glance:

• the flat projection: $CD = b\cos C = 5\cos 60^\circ = 2.5$;
• the height: $AD = b\sin C = 5\sin 60^\circ \approx 4.33$.

That leaves $DB = a - b\cos C = 8 - 2.5 = 5.5$. Right triangle $ADB$ now answers to Pythagoras:

$c^2 = (a - b\cos C)^2 + (b\sin C)^2$

Expand the squares: $c^2 = a^2 - 2ab\cos C + b^2\cos^2 C + b^2\sin^2 C$. Since $\cos^2 C + \sin^2 C = 1$, the tail folds into $b^2$. The find the value of 2abcosc term survives every rewrite, leaving

$c^2 = a^2 + b^2 - 2ab\cos C$

Numbers again: $5.5^2 + 4.33^2 = 30.25 + 18.75 = 49$, so $c = 7$ — the same 7 the head-check produced. One altitude, zero mystery: the find the value of 2abcosc term is a projection discount. That is also why the find the value of 2abcosc sign tracks the angle type. Every find the value of 2abcosc computation stands on this picture.

What if the angle goes back to $90^\circ$? Then $\cos 90^\circ = 0$, the correction term $2ab\cos C$ vanishes, and the formula collapses to $c^2 = a^2 + b^2$ — plain Pythagoras. Wentworth's classic nicknames the rule the Generalized Theorem of Pythagoras: the right triangle is the family member whose find the value of 2abcosc term is zero.

Below $90^\circ$ the cosine stays positive and $c$ comes out shorter; above $90^\circ$ it turns negative and $c$ stretches past $\sqrt{a^2+b^2}$. Old notebooks file this argument under the cos theorem proof — one altitude does all the work. It shows exactly where the find the value of 2abcosc term lives. That is why find the value of 2abcosc questions make sense from sides alone.

Many worksheets hand you three sides and ask for exactly one thing: find the value of $2ab\cos C$. Typed into a search box, the same request reads find the value of 2abcosc. No cosine table needed — the find the value of 2abcosc move is one rearrangement.

Loney's 1893 Plane Trigonometry, Article 164, compresses it — the original find the value of 2abcosc citation:

$c^2 = a^2 + b^2 - 2ab\cos C, \quad \text{i.e.} \quad 2ab\cos C = a^2 + b^2 - c^2$

Everything a find the value of 2abcosc question wants sits in that one line.

One sentence: the two adjacent squares summed, minus the opposite square, is $2ab\cos C$. That is the find the value of 2abcosc rule, complete. Every find the value of 2abcosc shortcut ever taught is this sentence in symbols.

Smallest example — the classic find the value of 2abcosc drill: sides $a = 2$, $b = 3$, $c = 4$. Then

$2ab\cos C = 2^2 + 3^2 - 4^2 = 4 + 9 - 16 = -3$

Wait — negative? Stay calm; that is information, not error. Here $\cos C = \frac{-3}{2\times 2\times 3} = -\frac{1}{4}$, so $C$ is obtuse ($\approx 104.5^\circ$). Side 4 is the longest, so its opposite angle is the biggest. A negative cosine is the angle announcing its type — the fastest signal a find the value of 2abcosc answer ever sends. Reading that signal is half the point of a find the value of 2abcosc drill.

Keep the angle version too; it rides along free: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$. Three sides known and an angle wanted? Use that one. Every find the value of 2abcosc result secretly contains it. Divide your find the value of 2abcosc answer by 2ab and the cosine appears.

The request arrives under many spellings — law cosines, law of cosin, even law of cosines] with its stray bracket. The spaced co sine law query lands here as well, hunting the same find the value of 2abcosc target. Every spelling wants the same three squares, and every spelling earns the same find the value of 2abcosc answer.

Problem. A triangle has sides $a = 15$, $b = 36$, $c = 39$. Find the value of $2ab\cos C$, and state the size of angle $C$.

Solution. Run the rearranged line straight through:

$2ab\cos C = a^2 + b^2 - c^2 = 15^2 + 36^2 - 39^2 = 225 + 1296 - 1521 = 0$

A zero correction forces $\cos C = 0$, which forces $C = 90^\circ$. The triple 15-36-39 is Pythagorean ($15^2 + 36^2 = 39^2$); the longest side faces the right angle. The find the value of 2abcosc arithmetic spotted it with zero angle work. This is the find the value of 2abcosc drill at its friendliest. Not every find the value of 2abcosc result is this clean — enjoy it.

Answer. $2ab\cos C = 0$, and $C = 90^\circ$ — the tidiest find the value of 2abcosc result there is.

Loney's original example also asks for $\cos A$. By the angle version, $\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{1296 + 1521 - 225}{2\times 36\times 39} = \frac{2592}{2808} = \frac{12}{13}$. The numerator is exactly the value of $2bc\cos A$ — the same add-and-subtract routine, aimed one letter over. The find the value of 2abcosc method scales to every corner of a triangle. The find the value of 2abcosc arithmetic never changes.

Problem. A triangle has $a = 10$, $b = 11$, and included angle $C = 50^\circ$. Find the third side $c$, rounded to two decimals. A find the value of 2abcosc step comes first — that is the point of this example.

Solution. First, compute the entire correction term $2ab\cos C$ — pure find the value of 2abcosc work:

$2ab\cos C = 2\times 10\times 11\times\cos 50^\circ \approx 220\times 0.6428 \approx 141.41$

Second, feed it to the law of cosines — the find the value of 2abcosc value slots straight in:

$c^2 = 10^2 + 11^2 - 141.41 = 100 + 121 - 141.41 = 79.59$

Take the root: $c \approx 8.92$ — the find the value of 2abcosc detour already paid for it.

Size check: at $90^\circ$ the third side would measure $\sqrt{100 + 121} \approx 14.87$. The real angle is only $50^\circ$, so the sides lean closer and the third side must land shorter. Since 8.92 sits below 14.87, the direction is right. The find the value of 2abcosc first, square-root second routine polices itself. Run the find the value of 2abcosc step, take the root, sanity-check the size — the full find the value of 2abcosc workflow.

Answer. $c \approx 8.92$. (Adapted from Granville §59, Exercises, no. 8 — the find the value of 2abcosc logic, 1908 edition.)

Problem. Return to the opening scene: $CA = 11.5$ miles, $CB = 9.4$ miles, $\angle C = 59.5^\circ$. Find $AB$, rounded to one decimal — the find the value of 2abcosc step unlocks it.

Solution. Set $a = BC = 9.4$ and $b = CA = 11.5$. The wanted $AB$ is exactly $c$ — the side opposite angle $C$, the slot the find the value of 2abcosc term adjusts.

Compute the correction first — the find the value of 2abcosc step, surveyor style:

$2ab\cos C = 2\times 9.4\times 11.5\times\cos 59.5^\circ \approx 216.2\times 0.5075 \approx 109.73$

Then substitute:

$c^2 = 9.4^2 + 11.5^2 - 109.73 = 88.36 + 132.25 - 109.73 = 110.88$

Square root: $c \approx 10.5$ miles.

The mountain blocked the tape, not the arithmetic: two walkable legs plus one angle lock the third side down. The 1908 book, working from five-place tables (printed to five decimal places), printed 10.6 miles; a modern calculator gives 10.5. One find the value of 2abcosc computation carried the whole survey — that is the find the value of 2abcosc payoff in the wild.

Answer. $AB \approx 10.5$ miles — bought by one find the value of 2abcosc step and one square root.

1. Mismatching a side with its opposite angle. The side you want sits alone on the left of the equation; the other two sides feed the cosine. $2ab$ must travel with $\cos C$ — the angle facing $c$ is $C$, by naming convention. Writing $c^2 = a^2 + b^2 - 2ab\cos A$ is the classic accident, and it wrecks an otherwise perfect find the value of 2abcosc answer. Every find the value of 2abcosc setup depends on that pairing.

2. Dropping the negative sign on obtuse angles. Past $90^\circ$, $\cos C < 0$, so $-2ab\cos C$ turns into a positive addition and the third side grows longer. When your $2ab\cos C$ lands negative, think first: the angle at $C$ is obtuse. Do not rush to flip signs — a negative find the value of 2abcosc result is often the correct one.

3. Skipping the triangle inequality on three given sides. The two shorter sides must add to more than the longest side. Otherwise no triangle exists, and no find the value of 2abcosc computation can rescue it.

4. Hunting the big angle with the sine rule. The sine rule is two-faced: $95^\circ$ and $85^\circ$ share one sine, and the calculator reports only the smaller. With three sides in hand, let the cosine rule take the largest angle first. The remaining angles are then safe for sines — and your find the value of 2abcosc work stays unambiguous.

5. Leaving the calculator in radian mode. Before evaluating $\cos 50^\circ$, confirm the calculator reads DEG. In radian mode $\cos 50 \approx 0.965$, one wrong step sinks the whole find the value of 2abcosc computation — and every step after it inherits the error.